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4 years ago

10

Scientific notation of 86400

2 answers:

Diano4ka-milaya [45]4 years ago

4 0

86400 scientific notation is equal to 8.64×10^4 . it also write as 8.64×10000

Mandarinka [93]4 years ago

3 0

8.64×10^4

this is 86400 in scientific notation

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3 years ago

ASE (National Institute for Automotive Service Excellence) certified professionals must be retested every _______ years to keep

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3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

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Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

3 years ago

Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons a

dsp73

Answer:

$E=576.5V/m$

Explanation:

From the question we are told that:

Length $l=56.0cm=0.56m$

Distance apart $d=7.0mm=0.007m$

Electron Transferred $n=10^{-10}$

Therefore

Total Charge

Since Charge on each electron is

$e=1.602*10^{-19}$

Therefore

$T=1.602*10^{-19} *10^{10}$

$T=1.602*10^{-9}$

Generally the equation for Charge density is mathematically given by

$\rho=T/A$

Where

Area

$A=0.56*0.56$

$A=0.3136$

Therefore

$\rho=1.602*10^{-9}/0.3136$

$\rho=5.10*10^{-9}$

Generally the equation for Electric Field in the capacitor is mathematically given by

$E=\frac{\rho}{e_0}$

$E=\frac{5.10*10^{-9}}{8.85x10{-12}}$

$E=576.5V/m$

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3 years ago

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kirza4 [7]

Thermal is the wasted energy in a charger

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3 years ago